Find The Highest Altitude - Leetcode Solution

Problem Statement: You are given an array of integers 'gain'. The ith integer represents the gain or loss of altitude for a person in the ith hike. Your task is to find the highest altitude achieved by the person during the hike.

Solution: To solve this problem, we need to iterate through the array 'gain' and keep track of the current altitude. We also need to keep track of the highest altitude achieved so far. To do this, we can use a variable 'altitude' to keep track of the current altitude and another variable 'max_altitude' to keep track of the highest altitude achieved so far.

We can start by initializing both 'altitude' and 'max_altitude' to 0. Then, we can iterate through the array 'gain' and for each integer in the array, we can add it to the current altitude. After adding the gain or loss to the current altitude, we can check if the current altitude is greater than the maximum altitude achieved so far. If it is, we can update the 'max_altitude' variable with the new value.

Once we have iterated through all the elements in the 'gain' array, we will have the highest altitude achieved by the person during the hike in the 'max_altitude' variable. We can then return the 'max_altitude' variable as the answer.

Pseudo Code:

alt = 0 //initialize the current altitude to 0 max_alt = 0 //initialize the maximum altitude achieved so far to 0

for each gain in the array 'gains': alt += gain //add the gain or loss to the current altitude if alt > max_alt: //check if the current altitude is greater than the maximum altitude achieved so far max_alt = alt //if it is, update the maximum altitude achieved so far

return max_alt //return the maximum altitude achieved

Time Complexity: O(n), where n is the length of the 'gain' array. We need to iterate through the array once to calculate the maximum altitude achieved.

Space Complexity: O(1). We are using only two variables 'altitude' and 'max_altitude' to keep track of the current altitude and the maximum altitude achieved so far. Hence, the space complexity is constant.

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